Question: Compute
\[\prod_{k = 1}^{12} \prod_{j = 1}^{10} (e^{2 \pi ji/11} - e^{2 \pi ki/13}).\]
Explanation: Let
\[P(x) = \prod_{k = 1}^{12} (x - e^{2 \pi ki/13}).\]The roots of this polynomial are $e^{2 \pi ki/13}$ for $1 \le k \le 12.$  They are also roots of $x^{13} - 1 = (x - 1)(x^{12} + x^{11} + x^{10} + \dots + x^2 + x + 1).$  Thus,
\[P(x) = x^{12} + x^{11} + x^{10} + \dots + x^2 + x + 1.\]Now, $e^{2 \pi ji/11},$ for $1 \le j \le 10,$ is a root of $x^{11} - 1 = (x - 1)(x^{10} + x^9 + x^8 + \dots + x^2 + x + 1),$ so $e^{2 \pi ji/11}$ is a root
of
\[x^{10} + x^9 + x^8 + \dots + x^2 + x + 1.\]So, if $x = e^{2 \pi ji/11},$ then
\begin{align*}
P(x) &= x^{12} + x^{11} + x^{10} + \dots + x^2 + x + 1 \\
&= x^2 (x^{10} + x^9 + x^8 + \dots + x^2 + x + 1) + x + 1 \\
&= x + 1.
\end{align*}Hence,
\begin{align*}
\prod_{k = 1}^{12} \prod_{j = 1}^{10} (e^{2 \pi ji/11} - e^{2 \pi ki/13}) &= \prod_{j = 1}^{10} P(e^{2 \pi ji/11}) \\
&= \prod_{j = 1}^{10} (e^{2 \pi ji/11} + 1).
\end{align*}By similar reasoning,
\[Q(x) = \prod_{j = 1}^{10} (x - e^{2 \pi ji/11}) = x^{10} + x^9 + x^8 + \dots + x^2 + x + 1,\]so
\begin{align*}
\prod_{j = 1}^{10} (e^{2 \pi ji/11} + 1) &= \prod_{j = 1}^{10} (-1 - e^{2 \pi ji/11}) \\
&= Q(-1) \\
&= \boxed{1}.
\end{align*}